equation 4.1 INTRODUCTION

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1 4 The momentum equation 4.1 INTRODUCTION It is often important to determine the force produced on a solid body by fluid flowing steadily over or through it. For example, there is the force exerted on a solid surface by a jet of fluid impinging on it; there are also the aerodynamic forces (lift and drag) on an aircraft wing, the force on a pipe-bend caused by the fluid flowing within it, the thrust on a propeller and so on. All these forces are associated with a change in the momentum of the fluid. The magnitude of such a force is determined essentially by Newton s Second Law. However, the law usually needs to be expressed in a form particularly suited to the steady flow of a fluid: this form is commonly known as the steady-flow momentum equation and may be applied to the whole bulk of fluid within a prescribed space. Only forces acting at the boundaries of this fluid concern us; any force within this fluid is involved only as one half of an action-and-reaction pair and so does not affect the overall behaviour. Moreover, the fluid may be compressible or incompressible, and the flow with or without friction. 4.2 THE MOMENTUM EQUATION FOR STEADY FLOW In its most general form, Newton s Second Law states that the net force acting on a body in any fixed direction is equal to the rate of increase of momentum of the body in that direction. Since force and momentum are both vector quantities it is essential to specify the direction. Where we are concerned with a collection of bodies (which we shall here term a system) the law may be applied (for a given direction) to each body individually. If the resulting equations are added, the total force in the given fixed direction corresponds to the net force acting in that direction at the boundaries of the system. Only these external, boundary forces are involved because any internal forces between the separate bodies occur in pairs of action and reaction and therefore cancel in the total. For a fluid, which is continuum of particles, the same result applies: the net force in any fixed direction on a certain defined amount of fluid equals the total rate of increase of momentum of that fluid in that direction.

2 The momentum equation for steady flow 135 Fig. 4.1 Our aim now is to derive a relation by which force may be related to the fluid within a given space. We begin by applying Newton s Second Law to a small element in a stream-tube (shown in Fig. 4.1) The flow is steady and so the stream-tube remains stationary with respect to the fixed coordinate axes. The cross-section of this stream-tube is sufficiently small for the velocity to be considered uniform over the plane AB and over the plane CD. After a short interval of time δt the fluid that formerly occupied the space ABCD will have moved forward to occupy the space A B C D. In general, its momentum changes during this short time interval. If u x represents the component of velocity in the x direction then the element (of mass δm) has a component of momentum in the x direction equal to u x δm. The total x-momentum of the fluid in the space ABCD at the beginning of the time interval δt is therefore ABCD u x δm The same fluid at a time δt later will have a total x-momentum u x δm A B C D The last expression may be expanded as ABCD u x δm ABB A u x δm + DCC D u x δm The net increase of x-momentum during the time interval δt is therefore ( A B C D u x δm ) after δt ( ABCD u x δm ) before δt

3 136 The momentum equation ( = u x δm ) u x δm + u x δm ABCD ABB A DCC D ( ) u x δm ABCD before δt after δt ( ) = u x δm u x δm DCC D ABB A after δt since, as the flow is assumed steady, ( u x δm ) is the same after δt as ABCD before δt. Thus, during the time intereval δt, the increase of x-momentum of the batch of fluid considered is equal to the x-momentum leaving the stream-tube in that time minus the x-momentum entering in that time: ( ) ( ) u x δm u x δm DCC D ABB A For a very small value of δt the distances AA, BB are very small, so the values of u x, for all the particles in the space ABB A are substantially the same. Similarly, all particles in the space DCC D have substantially the same value of u x, although this may differ considerably from the value for particles in ABB A. The u x terms may consequently be taken outside the summations. Therefore the increase of x-momentum during the interval δt is (u )DCC D x δm (u ) x δm (4.1) ABB A Now ( δm ) DCC D is the mass of fluid which has crossed the plane CD during the interval δt and so is expressed by ṁδt, where ṁ denotes the rate of mass flow. Since the flow is steady, ( δm ) ABB A also equals ṁδt. Thus expression 4.1 may be written ṁ(u x2 u x1 )δt, where suffix 1 refers to the inlet section of the stream-tube, suffix 2 to the outlet section. The rate of increase of x-momentum is obtained by dividing by δt, and the result, by Newton s Second Law, equals the net force F x on the fluid in the x direction F x = ṁ(u x2 u x1 ) (4.2) The corresponding force in the x direction exerted by the fluid on its surroundings is, by Newton s Third Law, F x. A similar analysis for the relation between force and rate of increase of momentum in the y direction gives F y = ṁ(u y2 u y1 ) (4.3) In steady flow ṁ is constant and so ṁ = ϱ 1 A 1 u 1 = ϱ 2 A 2 u 2 where ϱ represents the density of the fluid and A the cross-sectional area of the stream-tube (A being perpendicular to u). We have so far considered only a single stream-tube with a cross-sectional area so small that the velocity over each end face (AB, CD) may be considered uniform. Let us now consider a bundle of adjacent stream-tubes, each of cross-sectional area δa, which together carry all the flow being examined.

4 The momentum equation for steady flow 137 The velocity, in general, varies from one stream-tube to another. The space enclosing all these stream-tubes is often known as the control volume and it is to the boundaries of this volume that the external forces are applied. For one stream-tube the x-force is given by δf x = ṁ(u x2 u x1 ) = ϱ 2 δa 2 u 2 u x2 ϱ 1 δa 1 u 1 u x1 The total force in the x direction is therefore F x = df x = ϱ 2 u 2 u x2 da 2 ϱ 1 u 1 u x1 da 1 (4.4a) (The elements of area δa must everywhere be perpendicular to the velocities u.) Similarly F y = ϱ 2 u 2 u y2 da 2 ϱ 1 u 1 u y1 da 1 (4.4b) and F z = ϱ 2 u 2 u z2 da 2 ϱ 1 u 1 u z1 da 1 (4.4c) These equations are required whenever the force exerted on a flowing fluid has to be calculated. They express the fact that for steady flow the net force on the fluid in the control volume equals the net rate at which momentum flows out of the control volume, the force and the momentum having the same direction. It will be noticed that conditions only at inlet 1 and outlet 2 are involved. The details of the flow between positions 1 and 2 therefore do not concern us for this purpose. Such matters as friction between inlet and outlet, however, may alter the magnitudes of quantities at outlet. It will also be noticed that eqns 4.4 take account of variation of ϱ and so are just as applicable to the flow of compressible fluids as to the flow of incompressible ones. The integration of the terms on the right-hand side of the eqns 4.4 requires information about the velocity profile at sections 1 and 2. By judicious choice of the control volume, however, it is often possible to use sections 1 and 2 over which ϱ, u, u x and so on do not vary significantly, and then the equations reduce to one-dimensional forms such as: F x = ϱ 2 u 2 A 2 u x2 ϱ 1 u 1 A 1 u x1 = ṁ(u x2 u x1 ) It should never be forgotten, however, that this simplified form involves the assumption of uniform values of the quantities over the inlet and outlet cross-sections of the control volume: the validity of these assumptions should therefore always be checked. (See Section ) A further assumption is frequently involved in the calculation of F. A contribution to the total force acting at the boundaries of the control volume comes from the force due to the pressure of the fluid at a cross-section of the flow. If the streamlines at this cross-section are sensibly straight and parallel, the pressure over the section varies uniformly with depth as for a static fluid; in other words, p = p+ϱgz is constant. If, however, the streamlines are not straight and parallel, there are accelerations perpendicular to them and consequent variations of p. Ideally, then, the control volume should be so selected that at the sections where fluid enters or leaves it the

5 138 The momentum equation streamlines are sensibly straight and parallel and, for simplicity, the density and the velocity (in both magnitude and direction) should be uniform over the cross-section. Newton s Laws of Motion, we remember, are limited to describing motions with respect to coordinate axes that are not themselves accelerating. Consequently the momentum relations for fluids, being derived from these Laws, are subject to the same limitation. That is to say, the coordinate axes used must either be at rest or moving with uniform velocity in a straight line. Here we have developed relations only for steady flow in a stream-tube. More general expressions are beyond the scope of this book Momentum correction factor By methods analogous to those of Section it may be shown that where the velocity of a constant-density fluid is not uniform (although essentially parallel) over a cross-section, the true rate of momentum flow perpendicular to the cross-section is not ϱu 2 A but ϱ u 2 da = βϱu 2 A. Here u = (1/A) uda, the mean velocity over the cross-section, and β is the momentum correction factor. Hence β = 1 ( u ) 2 da A A u It should be noted that the velocity u must always be perpendicular to the element of area da. With constant ϱ, the value of β for the velocity distribution postulated in Section is 100/98 = 1.02 which for most purposes differs negligibly from unity. Disturbances upstream, however, may give a markedly higher value. For fully developed laminar flow in a circular pipe (see Section 6.2) β = 4/3. For a given velocity profile β is always less than α, the kinetic energy correction factor. 4.3 APPLICATIONS OF THE MOMENTUM EQUATION The force caused by a jet striking a surface When a steady jet strikes a solid surface it does not rebound from the surface as a rubber ball would rebound. Instead, a stream of fluid is formed which moves over the surface until the boundaries are reached, and the fluid then leaves the surface tangentially. (It is assumed that the surface is large compared with the cross-sectional area of the jet.) Consider a jet striking a large plane surface as shown in Fig A suitable control volume is that indicated by dotted lines on the diagram. If the x direction is taken perpendicular to the plane, the fluid, after passing over the surface, will have no component of velocity and therefore no momentum in the x direction. (It is true that the thickness of the stream changes as the fluid moves over the surface, but this change of thickness corresponds to a negligible movement in the x direction.) The rate at which x-momentum enters

6 Applications of the momentum equation 139 Fig. 4.2 the control volume is ϱ 1 u 1 u x1 da 1 = cos θ ϱ 1 u 2 1 da 1 and so the rate of increase of x-momentum is cos θ ϱ 1 u 2 1 da 1 and this equals the net force on the fluid in the x direction. If the fluid on the solid surface were stationary and at atmospheric pressure there would of course be a force between the fluid and the surface due simply to the static (atmospheric) pressure of the fluid. However, the change of fluid momentum is produced by a fluid-dynamic force additional to this static force. By regarding atmospheric pressure as zero we can determine the fluid-dynamic force directly. Since the pressure is atmospheric both where the fluid enters the control volume and where it leaves, the fluid-dynamic force on the fluid can be provided only by the solid surface (effects of gravity being neglected). The fluid-dynamic force exerted by the fluid on the surface is equal and opposite to this and is thus cos θ ϱ 1 u 2 1 da 1 in the x direction. If the jet has uniform density and velocity over its cross-section the integral becomes ϱ 1 u 2 1 cos θ da 1 = ϱ 1 Q 1 u 1 cos θ (where Q 1 is the volume flow rate at inlet). The rate at which y-momentum enters the control volume is equal to sin θ ϱ 1 u 2 1 da 1. For this component to undergo a change, a net force in the y direction would have to be applied to the fluid. Such a force, being parallel to the surface, would be a shear force exerted by the surface on the fluid. For an inviscid fluid moving over a smooth surface no shear force is possible, so the component sin θ ϱ 1 u 2 1 da 1 would be unchanged and equal to the rate at which y-momentum leaves the control volume. Except when θ = 0, the spreading of the jet over the surface is not symmetrical, and for a real fluid the rate at which y-momentum leaves the control volume differs from the rate at which it enters. In general, the force in the y direction may be calculated if the final velocity of the fluid is known. This, however, requires further experimental data. When the fluid flows over a curved surface, similar techniques of calculation may be used as the following example will show.

7 140 The momentum equation Example 4.1 A jet of water flows smoothly on to a stationary curved vane which turns it through 60. The initial jet is 50 mm in diameter, and the velocity, which is uniform, is 36 m s 1. As a result of friction, the velocity of the water leaving the surface is 30 m s 1. Neglecting gravity effects, calculate the hydrodynamic force on the vane. Solution Taking the x direction as parallel to the initial velocity (Fig. 4.3) and assuming that the final velocity is uniform, we have Force on fluid in x direction = Rate of increase of x-momentum = ϱqu 2 cos 60 ϱqu 1 { π = (1000 kg m 3 ) 4 (0.05)2 m 2 36 m s 1} (30 cos 60 m s 1 36 m s 1 ) = 1484 N Similarly, force on fluid in y direction = ϱqu 2 sin 60 0 { = 1000 π 4 (0.05)2 36 kg s 1} (30 sin 60 m s 1 ) = 1836 N Fig. 4.3 The resultant force on the fluid is therefore ( ) N = 2361 N acting in a direction arctan {1836/( 1484)} = to the x direction. Since the pressure is atmospheric both where the fluid enters the control volume and where it leaves, the force on the fluid can be provided only by the vane. The force exerted by the

8 Applications of the momentum equation 141 fluid on the vane is opposite to the force exerted by the vane on the fluid. Therefore the fluid-dynamic force F on the vane acts in the direction shown on the diagram. If the vane is moving with a uniform velocity in a straight line the problem is not essentially different. To meet the condition of steady flow (and only to this does the equation apply) coordinate axes moving with the vane must be selected. Therefore the velocities concerned in the calculation are velocities relative to these axes, that is, relative to the vane. The volume flow rate Q must also be measured relative to the vane. As a simple example we may suppose the vane to be moving at velocity c in the same direction as the jet. If c is greater than u 1, that is, if the vane is receding from the orifice faster than the fluid is, no fluid can act on the vane at all. If, however, c is less than u 1, the mass of fluid reaching the vane in unit time is given by ϱa(u 1 c) where A represents the cross-sectional area of the jet, and uniform jet velocity and density are assumed. (Use of the relative incoming velocity u 1 c may also be justified thus. In a time interval δt the vane moves a distance cδt, so the jet lengthens by the same amount; as the mass of fluid in the jet increases by ϱacδt the mass actually reaching the vane is only ϱau 1 δt ϱacδt that is, the rate at which the fluid reaches the vane is ϱa(u 1 c).) The direction of the exit edge of the vane corresponds to the direction of the velocity of the fluid there relative to the vane. The action of a stream of fluid on a single body moving in a straight line has little practical application. To make effective use of the principle a number of similar vanes may be mounted round the circumference of a wheel so that they are successively acted on by the fluid. In this case, the system of vanes as a whole is considered. No longer does the question arise of the jet lengthening so that not all the fluid from the orifice meets a vane; the entire mass flow rate ϱau 1, from the orifice is intercepted by the system of vanes. Such a device is known as a turbine, and we shall consider it further in Chapter Force caused by flow round a pipe-bend When the flow is confined within a pipe the static pressure may vary from point to point and forces due to differences of static pressure must be taken into account. Consider the pipe-bend illustrated in Fig. 4.4 in which not only the direction of flow but also the cross-sectional area is changed. The control volume selected is that bounded by the inner surface of the pipe and sections 1 and 2. For simplicity we here assume that the axis of the bend is in the horizontal plane: changes of elevation are thus negligible; moreover, the weights of the pipe and fluid act in a direction perpendicular to this plane and so do not affect the changes of momentum. We assume too that conditions at sections 1 and 2 are uniform and that the streamlines there are straight and parallel.

9 142 The momentum equation Fig. 4.4 If the mean pressure and cross-sectional area at section 1 are p 1 and A 1 respectively, the fluid adjacent to this cross-section exerts a force p 1 A 1 on the fluid in the control volume. Similarly, there is a force p 2 A 2 acting at section 2 on the fluid in the control volume. Let the pipe-bend exert a force F on the fluid, with components F x and F y, in the x and y directions indicated. The force F is the resultant of all forces acting over the inner surface of the bend. Then the total force in the x direction on the fluid in the control volume is p 1 A 1 p 2 A 2 cos θ + F x This total x-force must equal the rate of increase of x-momentum ϱq(u 2 cos θ u 1 ) Equating these two expressions enables F x to be calculated. Similarly, the total y-force acting on the fluid in the control volume is p 2 A 2 sin θ + F y = ϱq(u 2 sin θ 0) and F y may thus be determined. From the components F x and F y the magnitude and direction of the total force exerted by the bend on the fluid can readily be calculated. The force exerted by the fluid on the bend is equal and opposite to this. If the bend were empty (except for atmospheric air at rest) there would be a force exerted by the atmosphere on the inside surfaces of the bend. In practice we are concerned with the amount by which the force exerted by the moving fluid exceeds the force that would be exerted by a stationary atmosphere. Thus we use gauge values for the pressures p 1 and p 2 in the above equations. The force due to the atmospheric part of the pressure is counterbalanced by the atmosphere surrounding the bend: if absolute values were used for p 1 and p 2 separate account would have to be taken of the force, due to atmospheric pressure, on the outer surface. Where only one of the pressure p 1 and p 2 is included in the data of the problem, the other may be deduced from the energy equation. Particular care is needed in determining the signs of the various terms in the momentum equation. It is again emphasized that the principle used is

10 Applications of the momentum equation 143 that the resultant force on the fluid in a particular direction is equal to the rate of increase of momentum in that direction. The force on a bend tends to move it and a restraint must be applied if movement is to be prevented. In many cases the joints are sufficiently strong for that purpose, but for large pipes (e.g. those used in hydroelectric installations) large concrete anchorages are usually employed to keep the pipe-bends in place. The force F includes any contribution made by friction forces on the boundaries. Although it is not necessary to consider friction forces separately they do influence the final result, because they affect the relation between p 1 and p 2. Example 4.2 A45 reducing pipe-bend (in a horizontal plane) tapers from 600 mm diameter at inlet to 300 mm diameter at outlet (see Fig. 4.5). The gauge pressure at inlet is 140 kpa and the rate of flow of water through the bend is m 3 s 1. Neglecting friction, calculate the net resultant horizontal force exerted by the water on the bend. Solution Assuming uniform conditions with straight and parallel streamlines at inlet and outlet, we have: u 1 = 0.45 m3 s 1 1 π = m s 4 (0.6 m)2 u 2 = m3 s 1 1 π = 6.01 m s 4 (0.3 m)2 By the energy equation ) p 2 = p (u ϱ 2 1 u2 2 = Pa kg m 3 ( ) m 2 s 2 = Pa Fig. 4.5

11 144 The momentum equation In the x direction, force on water in control volume = p 1 A 1 p 2 A 2 cos 45 + F x = ϱq(u 2 cos 45 u 1 ) = Rate of increase of x-momentum where F x represents x-component of force exerted by bend on water. Therefore Pa π m Pa π m 2 cos 45 + F x = 1000 kg m m 3 s 1 (6.01 cos ) m s 1 that is ( ) N + F x = 1168 N whence F x = N. In the y direction, force on water in control volume = p 2 A 2 sin 45 + F y = ϱq(u 2 sin 45 0) = Rate of increase of y-momentum, whence F y = (60.1 sin 45 ) N π sin 45 N = 7960 N Therefore total net force exerted on water = ( ) N = N acting in direction arctan {7960/( )} = to the x direction. Force F exerted on bend is equal and opposite to this, that is, in the direction shown on Fig For a pipe-bend with a centre-line not entirely in the horizontal plane the weight of the fluid in the control volume contributes to the force causing the momentum change. It will be noted, however, that detailed information is not required about the shape of the bend or the conditions between the inlet and outlet sections Force at a nozzle and reaction of a jet As a special case of the foregoing we may consider the horizontal nozzle illustrated in Fig Assuming uniform conditions with streamlines straight and parallel at the sections 1 and 2 we have: Force exerted in the x direction on the fluid between planes 1 and 2 = p 1 A 1 p 2 A 2 + F x = ϱq(u 2 u 1 ) If a small jet issues from a reservoir large enough for the velocity within it to be negligible (except close to the orifice) then the velocity of the fluid is increased from zero in the reservoir to u at the vena contracta (see Fig. 4.7). Consequently the force exerted on the fluid to cause this change is ϱq(u 0) = ϱqc v (2gh). An equal and opposite reaction force is therefore exerted by the jet on the reservoir.

12 Applications of the momentum equation 145 Fig. 4.6 Fig. 4.7 The existence of the reaction may be explained in this way. At the vena contracta the pressure of the fluid is reduced to that of the surrounding atmosphere and there is also a smaller reduction of pressure in the neighbourhood of the orifice, where the velocity of the fluid becomes appreciable. On the opposite side of the reservoir, however, and at the same depth, the pressure is expressed by ϱgh and the difference of pressure between the two sides of the reservoir gives rise to the reaction force. Such a reaction force may be used to propel a craft aircraft, rocket, ship or submarine to which the nozzle is attached. The jet may be formed by the combustion of gases within the craft or by the pumping of fluid through it. For the steady motion of such a craft in a straight line the propelling force may be calculated from the momentum equation. For steady flow the reference axes must move with the craft, so all velocities are measured relative to the craft. If fluid (e.g. air) is taken in at the front of the craft with a uniform velocity c and spent fluid (e.g. air plus fuel) is ejected at the rear with a velocity u r then, for a control volume closely surrounding the craft, The net rate of increase of fluid momentum backwards (relative to the craft) is ϱu 2 r da 2 ϱc 2 da 1 (4.5) where A 1, A 2 represent the cross-sectional areas of the entry and exit orifices respectively. (In some jet-propelled boats the intake faces downwards in the bottom of the craft, rather than being at the front. This, however, does not affect the application of the momentum equation since, wherever the water is taken in, the rate of increase of momentum relative to the boat is ϱqc. Nevertheless, a slightly better efficiency can be expected with

13 146 The momentum equation a forward-facing inlet because the pressure there is increased as in a Pitot tube so the pump has to do less work to produce a given outlet jet velocity.) Equation 4.5 is restricted to a craft moving steadily in a straight line because Newton s Second Law is valid only for a non-accelerating set of reference axes. In practice the evaluation of the integrals in eqn 4.5 is not readily accomplished because the assumption of a uniform velocity particularly over the area A 2 is seldom justified. Moreover, the tail pipe is not infrequently of diverging form and thus the velocity of the fluid is not everywhere perpendicular to the cross-section. In a jet-propelled aircraft the spent gases are ejected to the surroundings at high velocity usually greater than the velocity of sound in the fluid. Consequently (as we shall see in Chapter 11) the pressure of the gases at discharge does not necessarily fall immediately to the ambient pressure. If the mean pressure p 2 at discharge is greater than the ambient pressure p a then a force (p 2 p a )A 2 contributes to the propulsion of the aircraft. The relation 4.5 represents the propulsive force exerted by the engine on the fluid in the backward direction. There is a corresponding forward force exerted by the fluid on the engine, and the total thrust available for propelling the aircraft at uniform velocity is therefore (p 2 p a )A 2 + ϱu 2 r da 2 ϱc 2 da 1 (4.6) It might appear from this expression that, to obtain a high value of the total thrust, a high value of p 2 is desirable. When the gases are not fully expanded (see Chapter 11), however, that is, when p 2 > p a, the exit velocity u r relative to the aircraft is reduced and the total thrust is in fact decreased. This is a matter about which the momentum equation itself gives no information and further principles must be drawn upon to decide the optimum design of a jet-propulsion unit. Rocket propulsion A rocket is driven forward by the reaction of its jet. The gases constituting the jet are produced by the combustion of a fuel and appropriate oxidant; no air is required, so a rocket can operate satisfactorily in a vacuum. The penalty of this independence of the atmosphere, however, is that a large quantity of oxidant has to be carried along with the rocket. At the start of a journey the fuel and oxidant together form a large proportion of the total load carried by the rocket. Work done in raising the fuel and oxidant to a great height before they are burnt is wasted. Therefore the most efficient use of the materials is achieved by accelerating the rocket to a high velocity in a short distance. It is this period during which the rocket is accelerating that is of principal interest. We note that the simple relation F = ma is not directly applicable here because, as fuel and oxidant are being consumed, the mass of the rocket is not constant. In examining the behaviour of an accelerating rocket particular care is needed in selecting the coordinate axes to which measurements of velocities are referred. We here consider our reference axes fixed to the earth and all velocities must be expressed with respect to these axes. We may not consider